University of Connecticut OpenCommons@UConn
Honors Scholar Theses Honors Scholar Program
Spring 5-1-2018 Harmonic Functions and Harmonic Measure David McDonald [email protected]
Follow this and additional works at: https://opencommons.uconn.edu/srhonors_theses Part of the Partial Differential Equations Commons
Recommended Citation McDonald, David, "Harmonic Functions and Harmonic Measure" (2018). Honors Scholar Theses. 558. https://opencommons.uconn.edu/srhonors_theses/558 Harmonic Functions and Harmonic Measure
David McDonald
B.S., Applied Mathematics
An Undergraduate Honors Thesis Submitted in Partial Fulfillment of the Requirements for the Degree of Bachelor of Science at the University of Connecticut
May 2018 Copyright by
David McDonald
May 2018
i APPROVAL PAGE
Bachelor of Science Honors Thesis
Harmonic Functions and Harmonic Measure
Presented by David McDonald, B.S. Applied Mathematics
Honors Major Advisor HONORS ADVISOR NAME
Honors Thesis Advisor THESIS ADVISOR NAME
University of Connecticut May 2018
ii ACKNOWLEDGMENTS
I would like to thank Professor Murat Akman for guiding me through a subject that, prior to this year, I was unfamiliar with but interested in. I also would like to thank Professor Keith Conrad for sharing the tex files and guiding me for the process of writing this honors thesis. I’d also like to thank the University of Connecticut for giving me the opportunity to write this thesis, as well as my high school calculus teacher Mrs. Zackeo for teaching me how to do math correctly. Finally I’d like to thank my friends and family for always being there to support me; especially Leo for making sure his thesis is less interesting than mine, Will for raving to SHM with me, Chris Pratt for encouraging me to follow my dreams, Alicia for her assistance in the researching of Ja-Fire, Jacob for always having so much more work than me, and Kelly for helping me continue my streak of never losing at trivia.
iii Harmonic Functions and Harmonic Measure
David McDonald, B.S. University of Connecticut, May 2018
ABSTRACT
The purpose of this thesis is to give a brief introduction to the field of harmonic measure. In order to do this we first introduce a few important properties of harmonic functions and show how to find a Green’s function for a given domain. Following this we calculate the harmonic measure for some easy cases and end by examining the connection between harmonic measure and Brownian motion.
iv Contents
Introduction...... 1
Ch. 1. Harmonic Functions 2 1.1 Mean Value Property...... 4 1.1.1 Mean Value Property for Volume...... 5 1.2 The Maximum Principle...... 6 1.3 The Poisson Integral and Green’s function...... 8 1.3.1 Green’s Function for the Upper Half Plane...... 15 1.3.2 Green’s Function for the Unit Ball ...... 16 1.3.3 Green’s Function for the First Quadrant...... 18 1.3.4 Poisson’s Formula for the Ball...... 19
Ch. 2. Harmonic Measure 26 2.1 Examples ...... 28 2.1.1 The Upper Half Plane ...... 28 2.1.2 The Unit Disk...... 30 2.2 Brownian Motion...... 32 2.2.1 Harmonic Measure in Brownian Motion...... 33 2.2.2 Multiply Connected Domains ...... 35
v Introduction
Harmonic functions are those which satisfy Laplace’s equation. They have a num- ber of convenient properties including mean value properties and the Maximum Prin- ciple that make them easy to work with. Understanding harmonic functions is the first step to calculating the harmonic measure. The next step is finding Green’s func- tions; in section 1.3 we’ll introduce the Dirichlet problem and show how to solve a Green’s function. To make it less abstract there are a few examples: the upper half plane, the unit ball, and the first quadrant. Then, using the Green’s function for the unit ball, Poisson’s formula for the Ball can be found. These pieces are all integral to understanding and being able to find the harmonic measure. In probability theory, the harmonic measure ω of a particle in some domain Ω is the probability that it will exit Ω within some E along ∂Ω. At the beginning of chapter two we’ll calculate the harmonic measure for both the ball of radius r and the upper half plane as well as prove its uniqueness and existence in all cases. Harmonic measure has had quite a few recent theorems published, including Makarov’s theorem which was quite a breakthrough at the time, so we’ll conclude by going over those.
1 Chapter 1
Harmonic Functions
Before diving into harmonic measure, it is important to first have some knowledge of what a harmonic function is and the properties that it has. A harmonic function is any function that is twice continuously differentiable and satisfies the Laplace partial
differential equation, that is, in Rn, n ≥ 2,
n X ∂2u ∂2u ∆u = u = + ... + = 0. xixi ∂x2 ∂x2 i=1 1 n
Here ∆ is known as the Laplacian and denotes the sum of the second partial deriva- tives for each respective variable. One fundamental property of harmonic functions is that they are invariant under both rotations and translations. That is to say, both rotations and translations of harmonic functions are also harmonic.
2−n 2 2 2 (2−n)/2 Example 1.0.1. We shall see that v(x) = |x| = (x1+x2 ...+xn) is harmonic
n 2 2 1/2 2 in R \{0} when n > 2 and v(x) = ln |x| = ln(x1 + x2) is harmonic in R \{0}.
2 Let us start with n > 2, in this case, for i = 1, 2, . . . , n, we have
2 2 (−n)/2 vxi = (2 − n)xi(x1 + ... + xn) and
2 2 (−n)/2 2 2 2 (−n−2)/2 vxixi = (2 − n)[(x1 + ... + xn) − nxi (x1 + ... + xn) ].
Now
n n X X 2 2 (−n)/2 2 2 2 (−n−2)/2 vxixi = (2 − n) [(x1 + ... + xn) − nxi (x1 + ... + xn) ] i=1 i=1 n −n (−n−2)/2 X 2 = (2 − n)[n|x| − n|x| xi ] i=1 = (2 − n)n(|x|−n − |x|−n) = 0.
Hence v is harmonic in Rn.
2 2 1/2 When n = 2, i.e when v(x) = ln(x1 + x2) we have
2 xi 1 2xi vxi = 2 2 and vxixi = 2 2 − 2 2 2 . x1 + x2 x1 + x2 (x1 + x2)
From this we get
2 2 X X 1 2xi v = [ − ] xixi x2 + x2 (x2 + x2)2 i=1 i=1 1 2 1 2 2 2 2 X = − u2 x2 + x2 (x2 + x2)2 xi 1 2 1 2 i=1 2 2 = 2 2 − 2 2 = 0. x1 + x2 x1 + x2
3 1.1 Mean Value Property
Another important property of harmonic functions is called the mean value property. The mean value property states that, given u harmonic on the closed ball centered at a with radius r, B¯(a, r), u is equal to the average of u over ∂B(a, r). To be more precise, we have Z u(a) = u(a + rζ)dσ(ζ) S
where S is the unit sphere and σ is the Borel probability measure on the unit sphere so that σ(S) = 1.
Proof. First assume that n > 2 and take B(a, r) = B and ∈ (0, 1). Then using Green’s identity, it follows that
Z Z (u∆v − v∆u)dV = (uDnv − vDnu)ds Ω ∂Ω
with {Ω = x ∈ Rn : < |x| < 1} and v(x) = |x|2−n. This results in
Z Z Z Z 1−n 2−n 0 = (2 − n) uds − (2 − n) uds − Dnuds − Dnuds (1.1.1) S S S S
Here Dn is the differentiation with respect to the outward unit normal, Dn(ξ) = h∇u(ξ), n(ξ)i. Now since u is harmonic and v ≡ 1 we can apply another form of Green’s identity, Z Dnuds = 0. ∂Ω
4 By the identity above, the last two terms of (1.1.1) are 0, so we are left with
Z Z uds = 1−n uds S S
which is just Z Z udσ = u(ζ)dσ(ζ). S S
This is what we are looking to prove with → 0 with u continuous at 0. For n = 2 the proof is exactly the same, with the exception that |x|2−n be replaced with log |x|.
1.1.1 Mean Value Property for Volume
A similar property exists with respect to the volume of the ball rather than the surface, and its proof utilizes the polar coordinates formula for integration on Rn,
1 Z Z ∞ Z fdV = rn−1 f(rζ)dσ(ζ)dr (1.1.2) nV (B) Rn 0 S
The mean value property for volume states that if u is harmonic on B¯(a, r), then u(a) is equal to the average of u over B(a, r), or
1 Z u(a) = udV . V (B(a, r)) B(a,r)
5 Proof. Once again take B(a, r) = B and apply equation (1.1.2) setting f to be u multiplied by the characteristic function ξ of B. Following this, simply apply the standard mean value Property to get the desired result.
From the mean value property we gain an important insight about the singularities of harmonic functions.
Corollary 1.1.1. Any zeros of a real-valued harmonic function are not isolated.
Proof. Take u to be real-valued and harmonic on the open connected set Ω with a ∈ Ω and u(a) = 0. Now consider the closed ball B¯(a, r) ⊂ Ω with r > 0; since the average of the harmonic function u over the boundary of the ball is 0, u must either be identically 0 or it must take both positive and negative values on ∂B(a, r). Either way, this implies that u has a zero on ∂B(a, r). Therefor since u must have a zero on the boundary of any ball r > 0 centered at a. Hence a is not an isolated zero of u.
1.2 The Maximum Principle
Another important property of Harmonic Functions which stems from the mean value Property is called the maximum principle. The maximum principle states that for any real valued harmonic function u on Ω, where Ω is connected, if u attains a maximum or minimum in Ω then u must be constant.
6 Proof. For some a ∈ Ω, Consider u(a) to be a point where u attains its maximum. Now choose r > 0 such that B¯(a, r) ⊂ Ω. If at any point of B(a, r), u were less than u(a), then, by continuity, the average of u over B(a, r) would be less than u(a), which contradicts the mean value property. Therefore u must be constant on B(a, r), so the set where u attains its maximum is open in Ω. However the set is also closed in Ω because of u’s continuity, so the set must be all of Ω. So u must be constant on Ω. A similar argument can be made if u attains a minimum in Ω.
This principle leads to the following corollaries about real-valued harmonic func- tions.
Corollary 1.2.1. Suppose Ω is bounded and u is a continuous real valued function on Ω¯ that is harmonic on Ω. Then u must attain its maximum and minimum values over Ω¯ on ∂Ω.
Therefore a harmonic function on a bounded domain is determined by its own boundary values. So for u and v continuous on Ω¯ and harmonic on Ω with bounded Ω. Provided u = v on ∂Ω, then u = v on Ω. This is only guaranteed for a bounded domain however, and can fail in the unbounded case. For Ω unbounded, or when u is not continuous on Ω¯ we have a different version of the maximum principle.
Corollary 1.2.2. For u real valued and harmonic on Ω, suppose
lim sup u(ak) ≤ M k→∞
is true for every sequence ak in Ω converging to either a point in ∂Ω or ∞. Then u ≤ M on Ω.
7 0 Proof. Let M = sup u(x): x ∈ Ω and consider a sequence bk in Ω such that u(bk) →
0 0 M . If there exists a subsequence of bk that converges to a point b ∈ Ω then u(b) = M , and, by the maximum principle, u must be constant on the component of Ω that contains b. Therefore there must be a sequence ak converging to a boundary point of
0 0 Ω or ∞ where u = M , so M ≤ M. If no subsequence of bk converges to a point in
Ω, then there must exist a subsequence of bk converging to either a boundary point of Ω or ∞. Regardless, M 0 ≤ M for this case as well.
Since Corollaries 0.3 and 0.4 only apply to real functions, we need another corollary about the maximum principle for complex functions.
Corollary 1.2.3. Let Ω be connected with u harmonic on Ω. If |u| attains a maximum on Ω then u must be constant.
Proof. Suppose |u| attains a maximum value M at a point a ∈ Ω. Choose λ ∈ C so that |λ| = 1 and λu(a) = M. Now the real part of the harmonic function u attains a maximum value M at a; and by the maximum principle, Reλu ≡ M on Ω. And since |λu| = |u| ≤ M, the imaginary part of the harmonic function u, Imλu is equivalent to 0. Therefore λu and u are constant on Ω.
1.3 The Poisson Integral and Green’s function
In this section we want to solve the following Dirichlet problem. Let Ω be an open and bounded subset of Rn
( −∆u = f x ∈ Ω ⊂ n R (1.3.1) u = g x ∈ ∂Ω.
8 We will use Green’s function to solve this problem and for this we shall first give a motivation. Consider the following problem: ( −∆ G(x, y) = δ y ∈ Ω y x (1.3.2) G(x, y) = 0 y ∈ ∂Ω
Our first aim is to find such G. To this end, formally we have,
Z u(x) = δxu(y)dy Ω Z = − ∆yG(x, y)u(y)dy Ω Z Z ∂G = h∇yG(x, y), ∇yu(y)idy − (x, y)u(y) dS(y) Ω Ω ∂ν Z Z ∂u Z ∂G = − G(x, y)∆yu(y)dy + G(x, y) (y) dS(y) − (x, y)u(y) dS(y) Ω ∂Ω ∂ν Ω ∂ν Z Z ∂G = G(x, y)f(y) dy − (x, y)g(y) dS(y). Ω Ω ∂ν where we have used −∆u = f in Ω, u = g on ∂Ω and G(x, y) = 0 on ∂Ω. Here ν is the outer unit normal to Ω. Since we also know that
−∆yΦ(y) = δ0
−∆yΦ(x − y) = δx where Φ is the fundamental solution of the Laplace equation,
( 1 − 2π ln |x| n = 2 Φ(x) = 1 n(n−2)α(n)|x|n−2 n ≥ 3. Unfortunately Φ(x − y) doesn’t satisfy the given boundary conditions, but it can
9 still be used to help find a solution for (1.0.4). Integrating
Z Φ(x − y)∆u(y) dy Ω by parts is a bit tricky because Φ(x − y) has a singularity at x = y. To take care of this, fixed Ω, let > 0 such that dist(x, ∂Ω) < , where dist(x, ∂Ω) is the distance between x and ∂Ω, so that B(x, ) ⊂ Ω. Suppose for the moment that u ∈ C2(Ω) and using Divergence theorem and integration by parts we get
Z Z ∂u Z Φ(x − y)∆u(y)dy = Φ(x − y) (y)dS(y) − h∇yΦ(y − x), ∇yu(y)i dy V ∂V ∂ν V Z ∂u Z ∂Φ(x − y) Z = Φ(x − y) (y)dS(y) − u(y)dS(y) + ∆yΦ(x − y)u(y)dy ∂V ∂ν ∂V ∂ν V (1.3.3)
where V = Ω \ B(x, ) and once again ν is the outer unit normal to ∂V. Since
∆yΦ(x − y) = 0 in V we have
Z Z ∂u Z ∂Φ(x − y) Φ(x − y)∆u(y)dy = Φ(x − y) (y)dS(y) − u(y)dS(y). V ∂V ∂ν ∂V ∂ν (1.3.4)
As → 0 V → Ω, we want to send → 0 in the above identity and obtain
Z Z ∂u Z ∂Φ(x − y) Φ(x − y)∆u(y)dy = Φ(x − y) (y)dS(y) − u(y)dS(y). Ω ∂Ω ∂ν ∂Ω ∂ν (1.3.5)
To this end, first observe on the left-hand side of (1.3.4) converges to left-hand side
10 of (1.3.5) Z Z lim Φ(x − y)∆u(y)dy = Φ(x − y)∆u(y)dy. →0+ V Ω
It remains to show that the first and the second integral in (1.3.4) converges to first and the second integral in (1.3.5) respectively
Z ∂u Z ∂u lim Φ(x − y) (y)dS(y) = Φ(x − y) (y)dS(y) (1.3.6) →0+ ∂V ∂ν ∂Ω ∂ν and
Z ∂Φ(x − y) Z ∂Φ(x − y) lim u(y)dS(y) = u(y)dS(y) + u(x). (1.3.7) →0+ ∂V ∂ν ∂Ω ∂ν
This will finish (1.3.5). Let us start with (1.3.6). Notice that ∂V = ∂Ω ∪ ∂B(x, ). Therefore,
Z ∂u Z ∂u Z ∂u Φ(x − y) (y)dS(y) = Φ(x − y) (y)dS(y) + Φ(x − y) (y)dS(y). ∂V ∂ν ∂Ω ∂ν ∂B(x,) ∂ν
To show (1.3.6), we need to show
Z ∂u | Φ(x − y) (y)dS(y)| → 0 ∂B(x,) ∂ν as → 0+. To this end, using definition of Φ we get (we just do this when n ≥ 3 and
11 n = 2 is the same argument)
Z ∂u 1 Z 1 ∞ | Φ(x − y) (y)dS(y)| ≤ |∇u|L (B(x,)) n−2 dS(y) ∂B(x,) ∂ν n(n − 2)α(n) ∂B(x,) |x − y| 1 Z ∞ = n−2 |∇u|L (B(x,)) dS(y) n(n − 2)α(n) ∂B(x,) = C which clearly goes to zero as → 0+. Hence this finishes the proof of (1.3.6). Now we turn to (1.3.7). Using ∂V = ∂Ω ∪ ∂B(x, ) and on ∂B(x, ), the inner unit normal
y−x at y ∈ ∂B(x, ) is ν = − |y−x| . Moreover,
1 y − x ∇ Φ(y − x) = − y nα(n) |y − x|n
Then
∂Φ(x − y) 1 y − x y − x 1 1 = h∇Φ(x − y), νi = − h , − i = . ∂ν nα(n) |y − x|n |y − x| nα(n) |y − x|n−1
Hence using these in (1.3.7) we get
Z ∂Φ(x − y) Z ∂Φ(x − y) Z ∂Φ(x − y) u(y)dS(y) = u(y)dS(y) + u(y)dS(y) ∂V ∂ν ∂Ω ∂ν ∂B(x,) ∂ν Z ∂Φ(x − y) 1 Z 1 = u(y)dS(y) + n−1 u(y)dS(y) ∂Ω ∂ν nα(n) ∂B(x,) |y − x| Z ∂Φ(x − y) 1 Z = u(y)dS(y) + n−1 u(y)dS(y). ∂Ω ∂ν nα(n) ∂B(x,)
12 If we let → 0+ and use the mean value property for u we get
1 Z n−1 u(y)dS(y) → u(x). nα(n) ∂B(x,)
This finishes the proof of (1.3.7). Combining these we get
Z ∂u ∂Φ Z u(x) = Φ(y − x) (y) − (y − x)u(y) dS(y) − Φ(y − x)∆u(y) dy. ∂Ω ∂v ∂v Ω (1.3.8)
∂u We still cannot use this to solve (1.0.3), however, because we do not know ∂ν on ∂Ω. Because of this, we have to introduce a corrector function, hx(y) which satisfies
( x ∆yh = 0 y ∈ Ω hx(y) = Φ(y − x) y ∈ ∂Ω.
Suppose we can find a smooth function hx to satisfy these conditions. Then,
Z Z Z x x ∂u x h (y)∆u(y) dy = h (y) dS(y) − ∇yh (y) · ∇yu(y) dy Ω ∂Ω ∂ν Ω Z ∂u = hx(y) (y) dS(y) ∂Ω ∂ν Z x Z ∂h x − (y)u(y) dS(y) + ∆yh (y)u(y) dy ∂Ω ∂ν Ω and since hx is a solution to the given boundary conditions,
Z ∂u ∂hx Z 0 = Φ(y − x) (y) − (y)u(y) dS(y) − hx∆u(y) dy. (1.3.9) ∂Ω ∂ν ∂ν Ω
13 Subtracting equation (1.0.5) from (1.0.6) and allowing
G(x, y) = Φ(y − x) − hx(y) results in
Z ∂G Z u(x) = − (x, y)u(y) dS(y) − G(x, y)∆u(y) dy. (1.3.10) ∂Ω ∂ν Ω
The function G is known as Green’s function for Ω, it is formally defined as
G(x, y) = Φ(y − x) − hx(y) x, y ∈ Ω, x 6= y.
Therefor, when u is a smooth solution of a problem of the type (1.3.1) and G(x, y) is Green’s function then
Z ∂G Z u(x) = − (x, y)g(y) dS(y) + G(x, y)f(y) dy. (1.3.11) ∂Ω ∂ν Ω
Let us state this as a theorem.
Theorem 1.3.1. Let f, g be continuous function and if u ∈ C2(Ω)¯ is a solution of
n −∆u = f x ∈ Ω ⊂ R (1.3.12) u = g x ∈ ∂Ω
then for x ∈ Ω one has
Z ∂G Z u(x) = − (x, y)g(y) dS(y) + G(x, y)f(y) dy. (1.3.13) ∂Ω ∂ν Ω
14 where G(x, y) is the Green’s function for Ω.
We have an immediate corollary of this theorem.
Corollary 1.3.2. If u is a harmonic function in Ω with u = g on ∂Ω then
Z ∂G u(x) = − (x, y)g(y) dS(y). ∂Ω ∂ν
1.3.1 Green’s Function for the Upper Half Plane
2 For our first example, we’ll look to find the Green’s function for R+. In order to do
x 2 this, we’ll need a corrector function h for each x ∈ R+ such that
( ∆ hx(y) = 0 y ∈ 2 y R+ (1.3.14) 2 hx(y) = Φ(y − x) y ∈ ∂R+ 2 First fix x ∈ R+. Since ∆yΦ(y − x) = 0 for all y 6= x we choose z∈ / Ω, so that
∆yΦ(y − z) = 0 for all y ∈ Ω. Now all that’s left is to choose z = z(x), z∈ / Ω such that Φ(y − z) = Φ(y − x) for y ∈ ∂Ω, and let hx(y) = Φ(y − z(x)), we’ve found a corrector function. Since, for this example, n = 2,
1 Φ(y − z) = − ln |y − z|. 2π
2 This tells us that Φ(y − z) is a function of |y − z|, and so for x = (x1, x2) ∈ R+, for
2 y ∈ ∂R+,
|y − x| = |(y1, 0) − (x1, x2)| = |(y1, 0) − (x1, −x2)| = |y − x˜|.
Wherex ˜ is the reflection of x in the R2 plane.
15 y
2 R+ x
x y
x˜
Figure 1.3.1: Location ofx ˜ in the plane
2 x This means that our corrector function for R+ is h (y) = Φ(y − x˜). Furthermore,
2 a Green’s function for R+ is
1 G(x, y) = Φ(y − x) − Φ(y − x˜) = − (ln |y − x| − ln |y − x˜|). 2π
A Green’s function for the upper half space in Rn is still G(x, y) = Φ(y −x)−Φ(y −x˜) where Φ(x) = Φ(y − x) − Φ(y − x˜) is the fundamental solution when n ≥ 3, however
instead of havingx ˜ ≡ (x1, −x2),x ˜ ≡ (x1, ...xn−1, −xn).
1.3.2 Green’s Function for the Unit Ball
Finding the Green’s function for the unit ball is a bit more difficult, let
2 2 2 B2(0, 1) ≡ {(x1, x2) ∈ R : x1 + x2 < 1},
16 2 and fix x ∈ B2(0, 1). Since we are again working in R ,
1 Φ(y − x) = − ln |y − x|. 2π
Like before, this means that Φ(y − x) is a function of |y − x|. Now we need to find hx(y) = Φ(y − x) for all y ∈ ∂B2(0, 1), or all y such that |y| = 1. So for these y,
|y − x|2 = (y − x) · (y − x)
= |y|2 − 2y · x + |x|2
= |x|2 − 2x · y + 1
= |x|2|y|2 − 2x · y + 1 2x · y 1 = |x|2 |y|2 − + |x|2 |x|2 x |x|2 = |x|2 |y|2 − 2y · + |x|2 |x|4 = |x|2|y − x∗|2
∗ x with x = |x|2 being known as the point dual to x. ∗ ∗ Also, because x is not in B2(0, 1), we know that Φ(|x|(y − x )) is harmonic for all y in Ω. This Φ(|x|(y − x∗)) is also our corrector function for the unit ball because
∗ Φ(|x|(y − x )) = Φ(y − x) for all y ∈ ∂B2(0, 1). Therefor the Green’s function for the unit ball in R2 is
1 G(x, y) = Φ(y − x) − Φ(|x|(y − x∗)) = − (ln |y − x| − ln ||x||y − x∗|). 2π
The Green’s function for the unit ball in Rn is still G(x, y) = Φ(y−x)−Φ(|x|(y−x∗)),
17 y
x∗ B(0, 1)
x x
Figure 1.3.2: Location of x∗ in the plane where Φ(x) = Φ(y − x) − Φ(|x|(y − x∗)) is the fundamental solution when n ≥ 3,
2 2 2 however rather than B2(0, 1) ≡ {(x1, x2) ∈ R : x1 + x2 < 1},
2 2 2 Bn(0, 1) ≡ {(x1, ..., xn): x1 + x2 + ... + xn = 1}
1.3.3 Green’s Function for the First Quadrant
We shall also find the Green’s function for the first quadrant {(x, y) ∈ R2, x > 0, y >
∗ 0}. Let x = (x1, −x2) if x = (x1, x2). In this case the Green’s function is
1 1 1 1 G(x, y) = − ln |x − y| + ln |x∗ − y| − ln |x + y| + ln |x∗ + y| 2π 2π 2π 2π
for the first quadrant.
18 y
2 R+
x = (x1, x2)
x y
x˜ = (x1, −x2)
Figure 1.3.3: Location ofx ˜ in the plane
1.3.4 Poisson’s Formula for the Ball
Consider the unit ball in Rn, we are looking for a solution to Laplace’s equation in
Bn(0, 1) with boundary conditions
( ∆u = 0 x ∈ B (0, 1) n (1.3.15) u = g x ∈ ∂Bn(0, 1).
We know from Corollary 1.3.2 that u has the form
Z ∂G u(x) = − (x, y)g(y) dS(y). ∂Bn(0,1) ∂ν
We also know from the previous example that the Green’s function for a unit ball on
Rn is G(x, y) = Φ(y − x) − Φ(|x|(y − x∗))
19 and that 1 1 Φ(y) = (1.3.16) nα(n) yn−2 for n ≥ 3. Working with (1.3.16), we see that
1 1 Φ(|x|(y − x∗)) = (1.3.17) nα(n) |x|(y − x∗)|n−2 1 = Φ(y − x∗). (1.3.18) |x|n−2
Beginning again with (1.3.16), we find
y ∇Φ(y) = − nα(n)|y|n and y − x ∇ Φ(y − x) = − (1.3.19) y nα(n)|y − x|n
Combining (1.3.18) with (1.3.19) we see that
1 y − x∗ ∇ Φ(|x|(y − x∗)) = − y |x|n−2 nα(n)|y − x∗|n y|x|2 − x = − nα(n)||x|(y − x∗)|n y|x|2 − x = − nα(n)|y − x|n
Since we are working with the unit ball, the unit normal vector to Bn(0, 1) is just
y ν = = y. |y|
20 Thus, we take the normal derivative of G(x, ·) on ∂Bn(0, 1),
∂G ∂Φ ∂Φ = (y − x) − (|x|(y − x∗)) ∂ν ∂ν ∂ν y − x |x|2 − x = − · y + · y nα(n)|y − x|n nα(n)|y − x|n ky|2 + x · y + |y|2|x|2 − x · y = nα(n)|y − x|n |y|2(|x|2 − 1) = nα(n)|y − x|n |x|2 − 1 = . nα(n)|y − x|n
Now utilizing the form of u(x) given by Corollary 1.3.2, we know that
1 − |x|2 Z g(y) u(x) = n dS(y). (1.3.20) nα(n) ∂Bn(0,1) |y − x|
This formula is a solution for the unit ball, so now we want to find a solution formula for the ball of radius r, this time with boundary conditions
( ∆u = 0 x ∈ B (0, r) n (1.3.21) u = g x ∈ ∂Bn(0, 1).
To do this, we take u to be a solution to (1.3.21), and defineu ˜(x) = u(rx),
21 g˜(x) = g(rx), andy ˜ = ry. Then use these in (1.3.20) to obtain
1 − |x|2 Z g˜(y) u˜(x) = n dS(y) nα(n) ∂Bn(0,1) |y − x| 1 − |x|2 Z g(˜y) = n dS(˜y) nα(n) ∂Bn(0,r) |y/r˜ − x| rn(1 − |x|2) Z g(˜y) = n dS(˜y) nα(n) ∂Bn(0,r) |y˜ − rx| r2 − |rx|2 Z g(y) = n dS(˜y) nα(n)r ∂Bn(0,r) |y˜ − rx|
So a solution for the given boundary conditions would be
r2 − |x|2 Z g(y) u(x) = n dS(y). (1.3.22) nα(n)r ∂Bn(0,r) |y − x|
This equation is known as Poisson’s formula for the ball. In order to show that (1.3.22) is a solution to Laplace’s equation and satisfies the boundary conditions given by (1.3.1), we need to make use of the following theorem.
Theorem 1.3.3. For g ∈ C(∂Bn(0, r)) and u as defined by Poisson’s formula for the ball, u satisfies
∞ 1. u ∈ C (Bn(0, r))
2. ∆u = 0 for x ∈ Bn(0, r)
3. for x ∈ Bn(0, r) and all x0 ∈ ∂Bn(0, r), limx→x0 u(x) = g(x0)
To prove (1.3.3) we first need to prove that Green’s functions are symmetric,
Lemma 1.3.4. For all x, y ∈ Ω, x 6= y
G(x, y) = G(y, x)
22 Proof. Begin by fixing x, y ∈ Ω, x 6= y, and let v(z) ≡ G(x, z) and w(z) ≡ G(y, z). We need to show that v(y) = w(x) in order to show that the functions are symmetric. Remember that Green’s function is
G(x, y) = Φ(y − x) − hx(y) where hx(y) satisfies boundary conditions
( x ∆yh = 0 y ∈ Ω hx(y) = Φ(y − x) y ∈ ∂Ω.
Therefore both G(x, z) = 0 and G(y, z) = 0. Since v and w are smooth everywhere except when z = x and z = y, respectively, we should define a new region V such that V = Ω − [B(x, ) ∪ B(y, )]. Since both v and w are smooth on this new region, we can integrate by parts,
Z Z Z ∂v ∆vw dz = − ∇v · ∇w dz + w dS(z) V V ∂V ∂ν Z Z ∂w Z ∂v = v∆w dz − v dS(z) + w dS(z) V ∂V ∂ν ∂V ∂ν
Since ∆w = ∆v = 0 the first term on the right and the term on the left both vanish
Z ∂w Z ∂v 0 = − v dS(z) + w dS(z) ∂V ∂ν ∂V ∂ν and thus, Z ∂w Z ∂v v dS(z) = w dS(z) ∂V ∂ν ∂V ∂ν
23 Since v = w = 0 on ∂Ω, we can rewrite this equality as
Z ∂v ∂w Z ∂w ∂v w − v dS(z) = v − w dS(z) (1.3.23) ∂B(x,) ∂ν ∂ν ∂B(y,) ∂ν ∂ν
Now we must show that as approaches 0, the left and right sides converge to v(x) and w(y) respectively. Beginning with the left side,
Z ∂v ∂w Z ∂v Z ∂w w − v dS(z) = w dS(z) − v dS(z) ∂B(x,) ∂ν ∂ν ∂B(x,) ∂ν ∂B(x,) ∂ν Z ∂Φ ∂hx = (z − x) − (z) w dS(z) ∂B(x,) ∂ν ∂ν Z ∂w − v dS(z) ∂B(x,) ∂ν Z ∂Φ Z ∂hx = (z − x)w(z) dS(z) − (z)w dS(z) ∂B(x,) ∂ν ∂B(x,) ∂ν Z ∂w − v dS(z) ∂B(x,) ∂ν
We will show that the last two terms converge to zero while the first term converges
∂w to w(x). Beginning with the final term, since w is smooth near x, we know that ∂ν is bounded near ∂B(x, ), this means that v(z) = Φ(z − x) − hx(z) and
Z Z ∂w n−1 v dS(z) ≤ C sup |v| dS(z) = C sup |v| = O() ∂B(x,) ∂ν ∂B(x,) ∂B(x,) ∂B(x,) which converges to 0 as → 0. The second term follows in a similar manner,
Z x Z ∂h n−1 (z)w dS(z) ≤ C dS(z) ≤ C . ∂B(x,) ∂ν ∂B(x,)
24 Which again converges to 0 as → 0. Finally for the first term,
Z ∂Φ 1 Z (z − x)w(z) dS(z) = n−1 w(z) dS(z) → w(x) as → 0. ∂B(x,) ∂ν nα(n) ∂B(x,)
Thus the left side converges to w(x); the right side converges to v(y) in a similar manner.
25 Chapter 2
Harmonic Measure
Harmonic measure stems from the Dirichlet problem and can be used to measure the growth of harmonic functions. The harmonic measure is denoted by ω(z, Ω,E), and measures the value of a harmonic function ω at a point z with a boundary limit of
1 at E and 0 at ∂Ω \ E, where Ω is a domain in the complex plane C consisting of finitely many closed curves called Jordan Curves. In this setting, the harmonic measure always exists, is always unique, and is a conformal invariant; its resolvability is highly dependent on the simplicity of the boundary of the domain, Ω.
Theorem 2.0.1. The harmonic measure, ω(z, Ω,E), of a harmonic function, u, always exists and is always unique.
Proof. Fix a Jordan domain Ω. Since for any Ω there exists a conformal mapping Φ:Ω → U where U is a domain bounded by finitely many circles. Now take F such that F ⊂ ∂Ω. Since F consists of finitely many circular arcs, its harmonic measure exists and is equal to the Poisson integral of F ’s characteristic function. We know from Carath´eodory’s Theorem that the extension of Φ to its boundaries is a
26 ψ − 1 (E ) E
Φ Conformal X
Ω
homeomorphism of the closures, so therefore we can use Φ to show that the harmonic Measure of Ω exists in all cases. Uniqueness for bounded Ω follows from the maximum principle, but for unbounded Ω we need to use an extension called Lindel¨of’s maximum principle. It states that for any domain Ω whose boundary is not a finite set, let u be a real valued harmonic function on Ω and take M > 0 such that
u(z) ≤ M for all z ∈ Ω.
Now take m to be a real valued constant such that
lim sup u(z) ≤ m z→ζ for all except finitely many points of ζ ∈ Ω. Then u(z) ≤ m for all z ∈ Ω. With this extension of the maximum principle we cover all cases for Ω bounded and unbounded, so the harmonic measure is always unique.
Remark 2.0.2. When domains are sufficiently smooth, then Green’s theorem implies
27 harmonic measure is given by the normal derivative of Green’s function times surface measure on the boundary. Therefore,to find a harmonic measure of a domain, the key thing is to find the gradient of Greens function.
2.1 Examples
2.1.1 The Upper Half Plane
For the first example, choose Ω to be the upper half of the complex plane, and E to be any interval [−T,T ] on the real axis. Given that the Poisson kernel for Ω in this case is 1 x2 P (x, y) = · . π x2 + y2
We can easily get this by using the Green’s function for the upper half space. That
2 is, for fixed x + iy ∈ R+, i.e., y > 0, from our earlier work we have
1 (ξ − z)2 + (η − y)2 G(x, y, ξ, η) = ln . 4π (ξ − x)2 + (η + y)2
2 Hence on the boundary of R+, i.e., η = 0 we find
∂G ∂G = h∇G, νi = − . ∂ν ∂η
Now if we do the algebra we get
∂G 1 x2 − = · . ∂η π x2 + y2
28 y
2 R+ x + iy
α
x −T T
Figure 2.1.1: Harmonic measure of upper half space for [−T,T ]
As we want to find the harmonic measure of [−T,T ] with fixed x + iy we need to solve the following Dirichlet problem
∆u = 0 in 2 , R+ u = 1 on y = 0 and − T ≤ x ≤ T, u = 0 on y = 0 and x ∈ R \ [−T,T ].
2 If we let ω(x + iy, R+, [−T,T ]) = u(x, y) we know that the solution is given he harmonic function is given by
1 Z ∞ y u(x, y) = ω(x + iy, 2 , [−T,T ]) = χ (t) dt R+ π [−T,T ] (x − t)2 + y2 −∞ (2.1.1) 1 Z T y = 2 2 dt. π −T (x − t) + y
This can be rewritten in terms of ω(x + iy, Ω,E) and simplified to yield
1 x + T 1 x − T ω(x + iy, Ω,E) = tan−1 − tan−1 π y π y
29 −1 −π π Where tan takes values from 2 to 2 . This implies that ω(x + iy, Ω,E) is equal α to π , where α is the angle subtended by the interval [−T,T ] at z. In only Cartesian coordinates, this is
∂G 1 y0 ω(z, Ω,E) = = − 2 2 . ∂n π (x − x0) + y0
2.1.2 The Unit Disk
For another example, rather than having E be a line segment, let it be the arc of a
circle in C from −i to i such that
2θ − α ω(z, D, E) = 2π
Here α is the central angle of the circle and θ is the angle subtended by E at the point z. Just like in the previous example we calculate the Poisson integral of the curve. Because our curve is part of a circle, the Poisson kernel is
1 − r2 P (θ) = r 1 − 2r cos(θ) + r2
And the full Poisson integral is
π 2 1 Z 2 1 − r ω(z, D, E) = 2 dt 2π −π 1 − 2r cos(λ − t) + r 2 (2.1.2) 2 π 1 − r Z 2 dt = 2 2 2π −π (1 + r ) − 2r cos(λ − t) + r 2
30 using the identity
Z dx 2 ra − b x =√ tan−1 tan a + b cos(x) a2 − b2 a + b 2 the integral simplifies to
1 1 + r λ + π/2 1 + r λ − π/2 ω(z, D, E) = tan−1 tan − tan−1 tan π 1 − r 2 1 − r 2
1 1 + r sin λ + 1 1 + r sin λ − 1 ω(z, D, E) = tan−1 · −tan−1 · π 1 − r cos λ 1 − r cos λ then applying the identity
a − b tan−1 a − tan−1 b = tan −1 1 + ab one can simplify the equation to
1 1 − r2 ω(z, D, E) = tan−1 . π 2r cos λ
This can also be written as
1 r2 − ρ2 ω(z, D, E) = 2πr r2 + ρ2 − 2rρ cos(θ − φ)
Where our starting point is defined by z = (ρ cos φ, ρ sin φ) and φ is the angle at center of point z with respect to the x axis.
31 2.2 Brownian Motion
A common problem that harmonic measure is used to solve is detecting an exit point
in Brownian motion. That is, given some particle in a domain Ω ∈ R that moves in accordance with Brownian motion, we are looking to find the point at which the particle first hits the boundary. In order to do this, we would place circular detectors of some radius r along ∂Ω which would detect when the particle hits.
Figure 2.2.1: Brownian motion inside ball and covering the boundary with balls
Given that these detectors have a cost of φ(r) we have to consider that, given a finite budget, we wouldn’t be able to find the exit point of certain Ω such as the unit disk, because the detectors would have to be so small that their radii sum to infinity.
However if φ(r)/r → 0 as r → 0 then covering the boundary with nk balls of radius
1/nk would result in us having a finite cost. Now consider ∂Ω to be the von Koch snowflake. It would take 4n balls of size 3−n to cover the boundary, so we can accomplish this on a finite budget. Though it would seem more expensive, it is actually cheaper to detect exit points on this fractal because all points of the boundary are not equally likely to be exit points like they were on the disk. We do this using the cost function φ(r) = r.
32 Figure 2.2.2: Brownian motion when boundary is the von Koch snowflake
2.2.1 Harmonic Measure in Brownian Motion
Given z ∈ Ω ⊂ R2 and E ⊂ ∂Ω the harmonic measure of E in Ω with respect to z is the probability that a Brownian motion started at z exits Ω somewhere in E. For fixed E this is a harmonic function with values in [0,1]. The minimum principle tells us that if the function vanishes anywhere on Ω then it vanishes everywhere. Therefore sets where ω(z, E, Ω) = 0 are not dependent on z. So we’re looking for conditions that would force the harmonic measure to be zero. Two of these conditions is the F. and M. Riesz Theorem (2.2.1) and Dahlberg’s Theorem:
Theorem 2.2.1. If Ω ∈ R2 is a simply connected planar domain curve such that
H1(∂Ω) < ∞, then harmonic measure is mutually absolutely continuous with respect to arc length. That is,
for E ⊂ ∂Ω, ω(z, E, Ω) = 0 ⇐⇒ H1(E) = 0
Where H1 is the one dimensional Hausdorff Measure and is equal to the length of the curve.
Theorem 2.2.2 (Dahlberg, [7]). For a bounded Lipschitz’ domain Ω ∈ Rn then both (n-1)-dimensional Hausdorff measure and harmonic measure are mutually absolutely
33 continuous. E ∈ ∂Ω, ω(z, E, Ω) = 0 ⇐⇒ Hn−1(E) = 0
Another theorem helpful theorem here is Makarov’s Theorem.
Theorem 2.2.3 (Makarov, [3]). If Ω is simply connected, then dim(ω) = 1.
Makarov’s theorem, published in 1985, was a huge jump in the field. He proved an even more precise version in the same year. This more precise version is known as Makarov’s Law of the Iterated Logarithm (LIL) for ω.
Theorem 2.2.4 (Makarov, [3]). Let
r 1 1 φ (t) = t exp C log log log log . C t t
Then there is a value A so that ω HφA for every simply connected domain and a value B so that ω ⊥ HφB for some simply connected domain.
The LIL has a very useful reduction when applied in a Bloch Space. A Bloch Space B is the space of holomorphic functions defined on the unit disc in the complex plane such that
0 ||f||B = sup |f (z)|(1 − |z|) < ∞. z∈D
0 0 With the knowledge that || log f ||B ≤ 2 =⇒ f is conformal =⇒ || log f ||B ≤ 6, and using
0 0 diam(J) = diam(f(I)) ≈ |f (zI )|diam(I) ≈ |f (zI )|ω(J),
34 where I is an interval on T and J = f(I) ⊂ ∂Ω, we find that the LIL reduces to
|g(reiθ)| lim supq = O(||g||B) for a.e. θ when g ∈ B. (2.2.1) r→1 1 1 log 1−r log log log 1−r
2.2.2 Multiply Connected Domains
Up until now, we have only dealt with simply connected domains. For a multiply connected domain all of dim(∂Ω) could be less than one, and so we wouldn’t have
ω Hα for all α < 1. For the multiply connected domains, the equation
Z ∂G X log dω = C + G(z) (2.2.2) ∂ν ∂Ω z:∇G(z)=0 is useful. In (2.2.2) C is the Robin constant of ∂Ω and G is the Green’s function for
∂G Ω. Since G > 0 both sides are bounded below,and since dω = ∂ν ds dω has density bounded below compared to arclength and ω is part of a one dimensional set. If ∂Ω has “many, well separated” components then dim(ω) < 1.
35 Bibliography
[1] Sheldon Axler, Paul Bourdon, and Wade Ramey. Harmonic function theory, vol- ume 137 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1992.
[2] John B. Garnett and Donald E. Marshall. Harmonic Measure, volume 2 of New Mathematical Monographs. Cambridge University Press, Cambridge, 2008.
[3] Nikolai G. Makarov. On the distortion of boundary sets under conformal map- pings. Proc. London Math. Soc. (3), 51(2):369–384, 1985. 34
[4] Christopher J. Bishop. Book Review of Harmonic Measure. Volume 44, Number 2, pages 267-276 of the Bulletin of the American Mathematics Society, 2006.
[5] Kakutani, Shizuo. Two-dimensional Brownian motion and harmonic functions. Proc. Imp. Acad. 20 (1944), no. 10, 706–714. doi:10.3792/pia/1195572706. https://projecteuclid.org/euclid.pja/1195572706
[6] Steven G. Krantz The Theory and Practice of Conformal Geometry. 169-189, Dover Publications, 2016.
[7] Bj¨ornE. J. Dahlberg. On the absolute continuity of elliptic measures. Amer. J. Math., 108(5):1119–1138, 1986. 33
36 [8] Julie Levandosky, Green’s Functions Lecture Notes. Stanford University, Math 220b, 2003, https://web.stanford.edu/class/math220b/handouts/greensfcns.pdf
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